For merely the cyclic part, consider elements of the form $(abcd) \in S_4$. How many of them are there? Indeed, the answer seems to be just $6 \times 5 \times 4 \times 3 = 360$ by choosing $a,b,c,d$ in that order. However, noting from the comment that $(abcd) = (bcda) = (cdab) = (dabc)$ ( in the cycle representation, for example $(1234) = (2341)$ ) tells you that we must divide by $4$ to avoid repetition. This leads to $90$ cycles.
However, when we are looking at the generated cyclic group, indeed note that $(abcd)$ and $(adcb)$ generate the same group (because if $x$ is of order $4$ then $x^3 = x^{-1}$ is also of order $4$ and hence generates the same group. However, $x^2$ is not, so we will have to be careful there).
Also, if two elements generate the same subgroup of order $4$ then either they are the same or inverses. Therefore, the above analysis gives $45$ distinct groups of order $4$.
Now, for the other part, we consider groups generated by $(abcd)(ef)$ which would be of order $4$ but would not coincide with any of the previous groups since there is always an element which does not have any fixed points here.
Choosing $(abcd)(ef)$ happens again in $90$ ways, since if we choose $(abcd)$ then $(ef)$ gets fixed for us. Once again, going to the cyclic subgroup , we can pair $(abcd)(ef)$ with its inverse which is $(adcb)(ef)$ for each element of this kind, once again resulting in $45$ distinct groups of order $4$.
Totalling the above gives $90$ distinct cyclic groups of order $4$, more than what is required to answer the question.
Apart from this if we choose to look at groups generated by two transpositions $(ab) \neq (cd)$, which are distinct from the previous ones since every element here has order at most two , then $(ab)$ has $15$ choices and $(cd)$ has $6$ choices, after which you remove their order of picking to get $45$ further choices. This leads to at least $135$ subgroups of order $4$. You can try to check if some others are there, my guess would be no.
Here is the solution(just the comments above in one answer):
We know, both $A_4$ and $V_4$ are characteristic subgroups.
First I show, that $A_4$ is fully invariant. We know that the order of the image divides the order of the preimage. As $A_4$ is generated by all elements of order 3, the image of these generators must have order 3 or 1. In both cases it is in $A_4$.
For showing that $V_4$ is fully-invariant, we look at an endomorphism which is not an automorphism. It follows that the kernel is non-trivial, as it is normal it must be $S_4$, $A_4$, $V_4$. In all cases $V_4$ is contained in the kernel, though mapped to $e \in V_4$.
Best Answer
Conjugacy classes in the symmetric group correspond to cycle-types. Thus, for $H = \langle (1234)\rangle$, the conjugacy class of $(1234)$ is just $\{(1234),(1243),(1324),(1342),(1423),(1432)\}$. We see that the cycles $(1234),(1432)$ generate the same group. Similarly, so do the cycles $(1243),(1342)$, and also the cycles $(1324),(1423)$.
Thus, the orbit of $H = \langle (1234)\rangle$ are the three groups $H, \langle (1243)\rangle,\langle (1324)\rangle$.
By orbit-stabilizer, the stabilizer $S$ is thus a subgroup of index 3, hence order 8. Clearly $S$ contains $H$, so to describe $S$, it suffices to find an element of $S$ not in $H$.
Note that every element of $H$ commutes with itself. On the other hand, we know that $(1234)$ and $(1432)$ are conjugate, so they must be conjugate by an element of $S_4$ not in $H$. Finding this conjugating element comes down to realizing that given two permutations $\sigma,\tau$, if $\tau(a) = b$, then $(\sigma\tau\sigma^{-1})(\sigma(a)) = \sigma(b)$. Thus, the cycle notation of $\sigma\tau\sigma^{-1}$ is precisely just $\sigma$ applied to the cycle notation of $\tau$. From this we see that the transposition $(24)$ has the property that $(24)(1234)(42) = (1432)$. Thus, the stabilizer $S$ is the group generated by $(1234)$ and $(24)$.